sign in sign up
<<
Home , Angular frequency

AP “C” – Unit 4/Chapter 10 & 11 Notes – Yockers Rotational

Angular , , and - rotational motion – motion of a body that spins about an axis - axis of – the line about which rotation occurs - – occurs when any single point on an object travels in a circle around an axis of rotation →r – radius →s – arc length – distance moved along the circumference of a circle - angles can be measured in → – an angle whose arc length is equal to its radius arc length 360 →rad = radian(s); 1 rad = /length of radius when s=r; 1 rad   57.3 2 → the radian is a pure number with no dimensions (ratio of lengths) →in general, any angle, , measured in radians, is defined by the relation s   or s  r r →conversions  rad  deg 180 - angular , , describes how much an object has rotated →the angle through which a point, line, or body is rotated in a specified direction and about a specified axis → - s s s  s      f  0  f 0 f 0 r r r s   r - s is positive (+) when rotation is counterclockwise (ccw) - s is negative (-) when rotation is clockwise (cw) - angular speed, , describes of rotation →the rate at which a body rotates about an axis, usually expressed in rad/s (or s-1 because radians are not dimensional) →average angular speed = avg

  f 0 avg   unit is rad/s or revolutions/unit t t f  t0 - instantaneous angular speed  d   lim  t0 t dt - , , occurs when angular speed changes →the time rate of change of angular speed, expressed in rad/s/s or rad/s2 (or s-2) →average angular acceleration = avg

  f 0  avg   t t f  t0

- instantaneous angular acceleration  d   lim  t0 t dt - model - all points on a rotating rigid object have the same angular acceleration and angular speed →if not, the object’s shape would change (it would not be rigid) - similarities of angular and linear quantities linear angular displacement s  speed v  acceleration a  - direction of  and  →the direction of  is along the axis of rotation - if rotation is counterclockwise then the direction of  is out of the plane of the diagram - if rotation is clockwise then the direction of  is into the plane of the diagram  d →the direction of  follows from its vector definition of (along the same axis as ) dt - for rotation about a fixed axis →the direction of  is the same as  if the angular speed (the magnitude of ) is increasing in time →the direction of  is antiparallel to  if the angular speed is decreasing in time

Rotational : The Rigid Body Under Constant Angular Acceleration - for rotational motion about a fixed axis, the simplest accelerated motion to analyze is motion under constant angular acceleration →manipulating the equation for instantaneous angular acceleration results in

 f t d   dt   f  0 t  0 0 →manipulating the equation for instantaneous angular speed results in  t t f 1 2 d   dt  0 t dt   f 0 0t  t  0 0 2 0 →eliminating  results in 1  f 0  2 0  f t →and eliminating t results in 2 2  f  0  2 f 0 

Tangential and centripetal acceleration - objects in circular motion have a tangential speed - an object farther from the axis of rotation must travel at a higher tangential speed around the circular path, s, to travel the same angular displacement as would an object closer to the axis

- tangential speed ds d v   r dt dt v  r - tangential acceleration is tangent to the circular path →the instantaneous linear acceleration of an object directed along the tangent to the object’s circular path dv d a   r t dt dt

at  r - centripetal acceleration    v f  v0 →remember a  and that can be changed either by magnitude or t f  t0 direction; a change in either produces acceleration →for an object moving in a circular path with a constant speed, acceleration is due to a change in direction →centripetal acceleration (ac) – acceleration directed toward the center of a circular path - magnitude of ac is given by v2 a  t c r

- because tangential speed is related to the angular speed by vt  r , the centripetal acceleration can also be found using angular speed 2 ac  r - http://www.phy.ntnu.edu.tw/oldjava/circularMotion/circular3D_e.html - http://www.walter-fendt.de/ph14e/carousel.htm - tangential and centripetal are perpendicular →tangential component of acceleration is due to changing speed →centripetal component of acceleration is due to changing direction - total acceleration is found using the Pythagorean theorem 2 2 2 2 2 4 2 4 a  at  ac  r   r   r   - direction of the acceleration can be found using a   tan 1 c at

Moment of and Rotational Kinetic - if a rigid object is considered to be a collection of particles that rotates about a fixed axis with an angular speed

→each particle has some determined by its and tangential speed →if the mass of the ith particle is mi and its tangential speed is vi, the kinetic energy of the particle is 1 2 Ki  2 mivi →the total kinetic energy KR of the rotating object is

1 2 1 2 2 1  2  2 1 2 KR   Ki   2mivi  2 miri   2 miri  K  2 mv i i i  i  - the quantity in parentheses above is the of inertia, I, of the rigid object 2 I  miri i → is the rotational analog of mass →the greater the moment of inertia of a body, the greater its resistance to change in its angular speed - therefore, the kinetic energy of the rotating rigid object can be expressed as 1 2 KR  2 I - on the storage side of the continuity equation for energy (Unit 3 Notes), the kinetic energy term should now be considered the sum of the changes in both translational and rotational kinetic energy - for an extended continuous object, the moment of inertia can be calculated by dividing the object into many small elements 2 2 I  lim r mi  r dm m 0   i i - it is usually easier to calculate the moments of inertia in terms of the volume of the elements rather than their mass (especially if  is uniform) m   V I   r 2 dV - examples (moment of inertia problems)

Torque -  (tau) – SI derived unit is N·m - point mass vs. extended object →extended object – has a definite, finite size and shape →point mass – assuming all of an object’s mass is concentrated at its center of mass - rotational and translational motion can be separated →translational motion – motion along a path →rotational motion – the motion of a body that spins about an axis - net produces rotation →torque – a quantity that measures the ability of a to rotate an object about some axis (torque measures the effectiveness of a given force in producing rotation of a body about a specified axis) →torque is not : work requires motion and torque does not; work is a dot product (the product of two vectors results in a scalar quantity) whereas torque is a cross product (the product of two vectors results in a third vector quantity) - torque depends on a force and a lever arm →how easily an object rotates depends not only on how much force is applied but also on where the force is applied →lever arm – the ┴ distance from the axis of rotation to a line drawn along the direction of the force - torque also depends on the angle between a force and a lever arm → do not have to be ┴ to cause rotation    Fr(sin)    Fd

- two equal but opposite forces can produce a rotational acceleration if they do not act along the same line - if are equal and opposite, there will be no rotational acceleration →seesaw – momentary torque produced by pushing with legs

Rotational Equilibrium - equilibrium requires zero net force and zero net torque →if the net force on an object is zero, the object is in translational equilibrium →if the net torque on an object is zero, the object is in rotational equilibrium - the dependence of equilibrium on the absence of net torque is called the condition of equilibrium →the resultant torque acting on an object in rotational equilibrium is independent of where the axis is placed - an unknown force that acts along a line passing through this axis of rotation will not produce any torque - beginning a diagram by arbitrarily setting an axis where a force acts can eliminate an unknown in a problem - conditions for equilibrium Type of Equation Symbolic Eq. Meaning translational Fnet=0 Fnet on an object must be zero rotational net=0 net on an object must be zero - unstable equilibrium – center of mass is above support point - stable equilibrium – center of mass is below support point

The Rigid Object Under a Net Torque - only the tangential component of an applied force contributes to the torque of a rotating rigid object - the radial component of an applied force provides no torque because its line of is through the axis of rotation - the torque on each particle in a rigid rotating object

Fti  mi ati

ri Fti  ri mi ati 2  i  mi ri  i - the sum of the torque on all particles in a rigid rotating object 2  i  miri i i i

 2    miri   i  - recognizing moment of inertia in the second equation above   I

 F  ma - the net torque on a rigid body is proportional to the body’s angular acceleration

Work and Energy in Rotational Motion - the work done by external forces on an object will equal the change in rotational kinetic energy as long as energy is not stored by any other means - work done by a torque   dW  F ds  F sinr d

→the radial component of F does no work because it is perpendicular to the displacement of the point of application of the force →the above equation can be written as dW  d this is analogous to the work done in translational motion (the product of force and translational displacement) →combining this result with the rotational form of Newton’s second law allows us to express the torque as d d d d   I  I  I  I  which leads to  d  dW  I d dt d dt d →integrating the above gives the total work done by the torque   f f 1 2 1 2 W   d  I d  I f  I0  K R   2 2 0 0 - if a system consists of components that are both translating and rotating, the work-kinetic

energy theorem generalizes to W  K  K R - the rate at which work is being done by F on an object rotating about a fixed axis (instantaneous , P) dW d  dt dt dW P   dt P  Fv

Angular - for a rotating object (with no translational motion), each particle in the object is moving in a circular path, so momentum is associated with the motion of each particle - although the object has no linear momentum (why not?), a “quantity of motion” is associated with its rotation () - the instantaneous angular momentum, L, of the particle relative to the axis of rotation is defined by the vector product of its instantaneous position vector r and the instantaneous linear momentum p    L  r  p

→SI units of angular momentum are kg·m2/s →the sense of L is governed by the right-hand rule   →because p  mv , the magnitude of L is L  mvrsin - L is zero when r is parallel to p ( = 0° or 180°) - L is a maximum equal to mvr when r is perpendicular to p ( = 90° or 270°) - the net torque acting on a particle equals the time rate of change of the particle’s angular Momentum dp   r  F  r  dt    L  r  p dL d dp dr (product rule for differentiation; cross product is not commutative)  r  p  r    p dt dt dt dt →the last term on the right is zero because v=dr/dt is parallel to p, so dL dp  r  substituting from the the box above, dt dt dL dp   which is the rotational analog of F  dt dt - the total angular momentum L of a system of particles about some point is defined as the vector sum of the angular momenta of the individual particles

L  L1  L2  Ln  Li i - the time rate of change of the total angular momentum of the system equals the vector sum of all torques, including those associated with internal forces between particles and those associated with external forces - the total angular momentum can vary with time only if there is a net external torque on the system dLi d dLtot  ext     Li  i dt dt i dt - determining the rotational analog of p  mv

→each particle of mass mi in a rigid object moves in a circular path of radius ri, with a tangential speed vi

 2  L  miviri  mi riri  miri   I i i  i  L  I

Conservation of Angular Momentum - the total angular momentum of a system is conserved if the net torque acting on the system is zero dL   tot  0  ext dt

Ltot  con  Ltot,i  Ltot, f →the total energy, linear momentum, and angular momentum of an isolated system are all conserved →the angular momentum of an isolated system is conserved whether the system is a rigid object or not

Parallel axis theorem - the total kinetic energy of a rolling object of mass M and moment of inertia I is the combination of the rotational kinetic energy around the center of mass plus the translational kinetic energy of the center of mass 1 2 1 2 K  2 ICM  2 MvCM - the parallel axis theorem permits the expression of the equation above in terms of the moment of inertia Ip through any axis parallel to the axis through the center of mass of the object 2 I p  ICM  MD →D is the distance from the center-of-mass axis to the parallel axis →M is the total mass of the object →so, the moment of inertia around an axis passing through the contact point P between a rolling object and the surface would be expressed as 2 I p  I CM  MR

→using the equation for total kinetic energy of a rolling object (above) and expressing the center of mass’ translational speed in terms of angular speed gives 1 2 1 2 K  2 I CM  2 MvCM K  1 I  2  1 MR 2 2 2 CM 2 1 2 2 K  2 I CM  MR  1 2 K  2 I p →thus, the kinetic energy of the rolling object can be considered as equivalent to a purely rotational kinetic energy of the object rotating around its contact point →depending on the situation, the equation for total kinetic energy of a rolling object can also be written as 1 2 1 2 K  2 I CM  2 MvCM 2 1  vCM  1 2 K  2 I CM    2 MvCM  R 

1  I CM  2 K  2   M vCM  R 2  - examples in book!

Kepler’s Laws - First - all planets move in elliptical orbits with the Sun at one of the focal points →any object bound to another by a force that varies as 1/r2 will move in an elliptical orbit - Second - a line drawn from the sun to any planet sweeps out equal in equal time intervals

→conservation of angular momentum

Lp  La

mvp rp  mva ra

v p rp  va ra - Third - the square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun

→period of a (derive this!)  4 2  2   3 3 2 3 T   r  K S r T  r  GM S  →period of a planet’s orbit around the Sun - for elliptical orbits, r is replaced with the length of the semimajor axis, a  4 2  T 2 T 2   a3  K a3 and  K  2.97 1019s 2 / m3 which is nearly constant   S 3 S  GM S  a for all the solar system’s planets! →period of a satellite’s orbit around the Earth  4 2  T 2   a3  K a3 19 2 3   E and K E  KS  2.97 10 s / m  GM E  - Conservation concepts and orbits →escape velocity (conservation of energy)

Ki U g,i  K f U g, f  GM m  GM m 1 mv2   E    E 2 i    RE  rf  1 1  v 2  2GM    i E    RE rf  2GM v  escape R

AP Physics “C” – Unit 4/Chapter 12 Notes – Yockers Oscillatory Motion

Motion of a Particle Attached to a - Hooke’s law   Fs  kx

→Fs = force of a spring (may be referred to as a “restoring force”) →k = spring constant (reflects the stiffness of the spring) →x = displacement (x = 0 is the equilibrium position) - the direction of the restoring force (Fs) is such that the object is being either pushed or pulled toward the equilibrium position - what is the description of an object’s motion in a system governed by Hooke’s law? →Fs →v and p →a - occurs when the net force along the direction of motion is a Hooke’s law type of force – that is, when the net force is proportional to the displacement and in the opposite direction - related terms →amplitude (A) ≡ the magnitude of the maximum position of the object relative to its equilibrium position →period (T) ≡ the time required for one complete → (f) ≡ the number of () per second (also the inverse of T)

- acceleration of an object undergoing SHM as a function of its position   ma  kx  k  a   x m since the maximum value of x is ±A, the acceleration ranges over the values –kA/m to +kA/m, so we also know that an object is undergoing SHM if its acceleration is proportional to its displacement and is in the opposite direction to it - in the absence of , the motion will continue forever because the force exerted by the spring is conservative

Mathematical Representation of Simple Harmonic Motion - position, velocity, and acceleration dv d 2 x →remember that a   , so dt dt 2  d 2 x k  a    x dt 2 m →the ratio k is denoted (explained later in notes) with the symbol  2 which results in m k   2 m d 2 x    2 x dt 2 →deriving the above from an equation describing the position of a particle in simple harmonic motion where A, , and  are constants of the motion - A (the amplitude of the motion) is the maximum value of the position of the particle in either the positive or negative direction k -  is the angular frequency found by   and has units of radians m per second -  is the phase constant (phase angle) and is determined uniquely by the position and velocity of the particle at t = 0. →if the particle is at its maximum position A at t = 0, the phase constant is  = 0

- position xt  Acost  - velocity dx d v   A cost   Asint  dt dt - acceleration d 2 x d a   A sint    2 Acost    2 x dt 2 dt

- time →period (T) of the motion is the time interval required for a particle to go through one full cycle of its motion →the values of x and v at time t equal the values of x and v at time t + T →angular frequency is related to the period by noting that the phase increases by 2 radians in a time interval of T t T t   2 2 1 2 2 m 1 1 k T     2f  T   2 f    f T  k T 2 m - maximum values of speed and acceleration k k v  A  A and a   2 A  A max m max m - evaluating the constants of the motion

→the graphs above show that the phase of the velocity differs from the phase of the position by /2 rad or 90° - when x is a maximum or a minimum, the velocity is zero - when x is zero, the speed is a maximum →the phase of acceleration differs from the phase of position by  or 180° - when x is a maximum, a has a maximum value in the opposite direction → is determined by k and m, T, or f (see the equations above) →the constants A and  are evaluated from the initial conditions (the state of the oscillator at t=0) using the equations for position, velocity and/or acceleration - examples

Energy Considerations in Simple Harmonic Motion - kinetic energy of a simple 1 2 1 2 1 2 2 2 K  2 mv  2 mAsint   2 m A sin t  - of a simple harmonic oscillator 1 2 1 2 1 2 2 U  2 kx  2 kAcost   2 kA cos t  - total energy of a simple harmonic oscillator E  K U

1 2 2 2 1 2 2 E  2 m A sin t   2 kA cos t   k E  1 m A2 sin 2 t   1 kA2 cos 2 t   2 m 2 1 2 2 2 E  2 kA sin t   cos t   1 2 E  2 kA →the total energy of an isolated simple harmonic oscillator is a constant of the motion and proportional to the square of the amplitude - the total energy is equal to the maximum potential energy stored in the spring when x ± A - at these points, v = 0 and there is no K →at the equilibrium position, x = 0 and U = 0, so the total energy is all in the form of kinetic energy of the particle 1 2 1 2 Kmax  2 mvmax  2 kA →the above results are appropriate for the simplification model in which the spring is considered to be massless

- the conservation of kinetic energy for an isolated system can be used to obtain the velocity for an arbitrary position x of the particle 1 2 1 2 1 2 E  K U  2 mv  2 kx  2 kA k v   A2  x 2    A2  x 2 m

The Simple - the motion of a pendulum is not SHM even though it is periodic - if the angular displacement from equilibrium is small (less than about 10 degrees or 0.2 rad, so sin=) then a pendulum follows the general equation form of Hooke’s law, and it is justified to say there is SHM as long as the size of the object is small relative to the length of the string

→ is about the lowest point which is the equilibrium position →the motion occurs in a vertical plane and is driven by the gravitational force (restoring force) which is always in the opposite direction of displacement →using Newton’s second law to write the equation of motion in the tangential direction d 2s F  ma   ma sin  m t t g dt 2 →because s  L and L is constant, the above equation can be written as d 2 a   g sin dt 2 L →for small angles sin   , so the equation of motion becomes d 2 a   g  dt 2 L - from this equation we can conclude that the motion is approximately simple

harmonic for small amplitudes because it contains the expression ag L 2 (remember   ag L ) - using the equation for displacement for a particle-spring system  can be

written as   max cost  where max is the maximum angular position and the angular frequency for a pendulum is

ag L   which gives the period as T  2 L ag →L = fixed length of the pendulum →ag = the acceleration due to (not necessarily the Earth’s!) →period does not depend upon the mass of the bob →period does not depend upon amplitude - how can a time piece using a pendulum be adjusted if it runs fast/slow? - examples

The Physical Pendulum - if a hanging object (that cannot be modeled as a particle) oscillates about a fixed axis that does not pass through its center of mass, it must be treated as a physical, or compound, pendulum - for a physical pendulum it is necessary to use a rigid object under a net torque

→the torque about O is provided by the gravitational force, and its magnitude is magdsin →applying Newton’s second law for rotation   I d 2  ma d sin  I g dt 2 - the negative sign on the left indicates that the torque about O tends to decrease  (the gravitational force produces a restoring force) →using the small angle approximation d 2  ma d    g  2   dt  I 

-   max cost  where max is the maximum angular position 2 →with the substitution   mag d I , the motion of the object is approximately simple harmonic for small amplitudes because the equation will now take the form of d 2 x    2 x dt 2 →angular frequency and period for a physical pendulum ma d 2 I   g and T   2 I  mag d Note: d is the distance from the axis of rotation to the center of mass (CM) as shown in the diagram above